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In the Wikipedia article of mapping cones there’s the following:

This complex is called the cone in analogy to the mapping cone (topology) of a continuous map of topological spaces $\phi \colon X \rightarrow Y$: the complex of singular chains of the topological cone $\mathrm{cone}(\phi)$ is homotopy equivalent to the cone (in the chain-complex-sense) of the induced map of singular chains of $X$ to $Y$.

I would like to see an explicit homotopy equivalence. We clearly have a chain map from $C_{n-1}(X) \oplus C_n(Y)$ to $C_n(\mathrm{cone}(\phi))$ (for an element $\sigma \in C_{n-1}(X)$ triangulate the cone $\sigma * \{1\}$; for an element in $C_n(Y)$ just map it to the base of the mapping cone). I couldn’t think of maps going in the other direction. Could someone help?

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  • $\begingroup$ Interesting question. I don't think a map $|\Delta^n|\to C_\phi$ need induce maps $|\Delta^n|\to Y$ or $|\Delta^{n-1}|\to X$ in any direct way, there's probably some 'swindle'. $\endgroup$
    – FShrike
    Dec 23, 2022 at 12:04

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$\DeclareMathOperator{\C}{Sing}\DeclareMathOperator{\Cone}{Cone}$ Added later. Probably the correct thing is to say that $C_\varphi$ is a homotopy pushout (the other map being the trivial map from $X$ to a point) in spaces while $\Cone(f)$ is a homotopy pushout in chain complexes, and that $\C$ preserves homotopy pushouts up to homotopy. I thought this was true but reading this post leaves me doubting.

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